All categories

2 years ago

During circular motion, the force that is perpendicular to the velocity and toward the center of the circle is the ... ?

1 answer:

5 0

That's the "centripetal" force. It produces the centripetal acceleration
that pulls the object away from a straight path into a bent path.

You might be interested in

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0

2 years ago

As populations in North America grow, so does the need for wood to build houses, stores, and other buildings. Select the stateme

Irina18 [472]

question 1, A

Question 2, D

Question 3, B

hope this helps :)

8 0

2 years ago

HELP!!!!!

devlian [24]

Answer:

Explanation:

6 0

2 years ago

The radius of Earth is . The average Earth-Sun distance is . How many Earths would fit between Earth and the Sun if they are sep

Verdich [7]

Answer:

The radius of the earth is 6,371 km.

The average Earth-Sun distance is 152.09 million km

How many Earths would fit between Earth and the Sun if they are separated by their average distance? Approximately 11,936 Earths.

I didn't really understand the last part, but if you don't get a better answer please mark me as brainliest.

6 0

2 years ago

calculating power reach each scenario and then answer the questions scenario a 120 j of work is done in 6 seconds scenario b 160

Molodets [167]

Answer: scenario b and scenario c uses most power

Explanation:

Scenario a:

Work=120J

Time=8 seconds

Power=work ➗ time

Power=120 ➗ 8

Power=15

Power=15 watts

Scenario b:

Work=160J

Time=8 seconds

Power=work ➗ time

Power=160 ➗ 8

Power=20

Power =20 watts

Scenario c:

Work=200J

Time=10 seconds

Power= work ➗ time

Power=200 ➗ 10

Power=20

Power=20 watts

Scenario b and scenario c uses most power

4 0

2 years ago